Câu 13 :

\(\left(-\frac{1}{4}+\frac{5}{8}\right)+-\frac{3}{5}\)

\(=\frac{3}{8}-\frac{-3}{5}\)

\(=\frac{39}{40}\)

Câu 14 :

\(M=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)

\(=\frac{5}{9}.\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)\)

\(=\frac{5}{9}.1=\frac{5}{9}\)

Câu 15 :

\(E=\left(-\frac{3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(E=\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)

\(E=0\)

Câu 16 :

\(H=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{8}+\frac{1}{36}-\frac{5}{12}\right)\)

\(=\frac{7}{8}:\frac{-7}{24}=-3\)

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Câu 14:

M =  \(\frac{5}{9}\) \(.\) \(\frac{7}{13}\) \(+\) \(\frac{5}{9}\) \(.\) \(\frac{9}{13}\) \(-\) \(\frac{5}{9}\) \(.\) \(\frac{3}{13}\)

M =    \(\frac{5}{9}\)  \(.\) (   \(\frac{7}{13}\)\(+\) \(\frac{9}{13}\) \(-\) \(\frac{3}{13}\))

M =      \(\frac{5}{9}\)  \(.\)    \(1\)

M  =              \(\frac{5}{9}\)

3: 

\(=\dfrac{1}{7}\cdot\dfrac{3}{5}\cdot\dfrac{5}{6}\cdot\dfrac{5}{8}=\dfrac{1}{7}\cdot\dfrac{1}{2}\cdot\dfrac{5}{8}=\dfrac{5}{112}\)

4:

=>2/3:x=-2-1/3=-7/3

=>x=-2/3:7/3=-2/7

5:

AC=CB=12/2=6cm

IB=6/2=3cm

a: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{4}=\dfrac{y}{-5}=\dfrac{-3x+2y}{-12-10}=\dfrac{55}{-22}=\dfrac{-5}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{-20}{2}=-10\\y=\dfrac{25}{2}\end{matrix}\right.\)

b: Ta có: \(\dfrac{x}{y}=\dfrac{-7}{4}\)

nên \(\dfrac{x}{-7}=\dfrac{y}{4}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{-7}=\dfrac{y}{4}=\dfrac{4x-5y}{-28-20}=\dfrac{72}{-48}=\dfrac{-3}{2}\)

Do đó: \(\left\{{}\begin{matrix}x=\dfrac{21}{2}\\y=\dfrac{-12}{2}=-6\end{matrix}\right.\)

c) \(\dfrac{x}{-3}=\dfrac{y}{8}\)   

⇒\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{x^2}{-9}=\dfrac{y^2}{64}=-\dfrac{44}{\dfrac{5}{-9+64}}=-\dfrac{44}{\dfrac{5}{55}}=-484\)

Bài 1:

a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)

b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)

Bài 2:

 \(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)

Bài 3:

\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)

\(\dfrac{2}{5}\) x y : \(\dfrac{7}{4}\) = \(\dfrac{7}{8}\)

\(\dfrac{2}{5}\) x y = \(\dfrac{7}{8}\) x \(\dfrac{7}{4}\)

 \(\dfrac{2}{5}\) x y = \(\dfrac{49}{32}\)

         y = \(\dfrac{49}{32}\) : \(\dfrac{2}{5}\)

         y = \(\dfrac{245}{64}\)

2\(\dfrac{2}{5}\): y x 1\(\dfrac{1}{4}\) = 2\(\dfrac{3}{5}\)

\(\dfrac{12}{5}\): y x \(\dfrac{5}{4}\) = \(\dfrac{13}{5}\)

\(\dfrac{12}{5}\): y        = \(\dfrac{13}{5}\): \(\dfrac{5}{4}\)

 \(\dfrac{12}{5}\): y = \(\dfrac{52}{25}\)

        y = \(\dfrac{12}{5}\): \(\dfrac{52}{25}\)

        y = \(\dfrac{15}{13}\)

\(\dfrac{12}{5}\) - 1\(\dfrac{2}{5}\) \(\times\) y = 1\(\dfrac{1}{4}\)

 \(\dfrac{12}{5}\) - \(\dfrac{7}{5}\) \(\times\) y  = \(\dfrac{5}{4}\)

           \(\dfrac{7}{5}\) \(\times\) y  = \(\dfrac{12}{5}\) - \(\dfrac{5}{4}\)

            \(\dfrac{7}{5}\) \(\times\) y = \(\dfrac{23}{20}\)

                   y = \(\dfrac{23}{20}\) : \(\dfrac{7}{5}\)

                   y = \(\dfrac{23}{28}\)

Ta có : 10 + 11+ 12 + 13 + ... + x = 5106

=> 1 + 2 + 3 + ..... + x = 5106 + (1 + 2 + 3 + ..... + 9)

=> 1 + 2 + 3 + ..... + x = 5106 + 45

=> 1 + 2 + 3 + ...... + x = 5151

=> \(\frac{x\left(x+1\right)}{2}=5151\)

<=> \(x\left(x+1\right)=10302\)

<=> x(x + 1) = 101.102

=> x = 101